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By irrationals, $\color{blue}{\mathbb{I}}$, I mean the set $\color{blue}{\mathbb{R}\setminus \mathbb{Q}}$ and the set $\color{blue}{\mathbb{R^2}\setminus\mathbb{Q^2}}$.

My thought is no in both cases.

For the set $\color{blue}{\mathbb{R}\setminus \mathbb{Q}}$, the set $\color{blue}{\mathbb{I}}$ is the disjoint union of the negative and positive open rays starting at $0$ each intersecting the $\color{blue}{\mathbb{I}}$ (to get the two open sets in the subspace topology to form a separation).

A similar argument for the set $\color{blue}{\mathbb{R^2}\setminus \mathbb{Q^2}}$ by separating it by two open half planes along the $y$-axis.

Is this argument correct?

Eric Wofsey
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Jack
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    It depends on what you mean by irrationals in the plane. – paw88789 Dec 13 '15 at 01:02
  • Could you clarify your questions? It's not at all clear what you are asking. – mweiss Dec 13 '15 at 01:02
  • what the heck does "connected" mean? – fleablood Dec 13 '15 at 01:02
  • Every single one of your phrases: "irrationals in the plane" "connected" "positive ray" "open ray" "starting at 0" "intersecting the irrationals" "open sets" "subsace topology" need to be defined. I have no idea what any of those mean. – fleablood Dec 13 '15 at 01:05
  • My guess would be that irrationals in the plane means ${(x,y)\mid x,y\in \mathbb{R}\setminus \mathbb{Q} }$ – Eff Dec 13 '15 at 01:10
  • Actually it's an exercise in Munkres to show that what you previously called the irrationals in the plane form a path connected set. – Matt Samuel Dec 13 '15 at 01:17
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    @fleablood if you don't know what an open set is, you should avoid asking for clarification on questions about topology. – Matt Samuel Dec 13 '15 at 01:26

2 Answers2

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You're right for $\mathbb{R}\setminus\mathbb{Q}$. However, your construction doesn't work for $\mathbb{R}^2\setminus\mathbb{Q}^2$, because there are points on the $y$-axis that are in $\mathbb{R}^2\setminus\mathbb{Q}^2$ (namely, points of the form $(0,y)$ where $y$ is irrational). These points won't be in either of your open half-planes.

In fact, $\mathbb{R}^2\setminus\mathbb{Q}^2$ is actually connected; see this question.

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Eric Wofsey
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I think the confusion comes from that $\mathbb R^2 \setminus \mathbb Q^2$ is not the same thing than $(\mathbb R \setminus \mathbb Q)^2$.

In the first definition these are points of the plane which do not have both coordinates rationnal, thus there can be mixed coordinates, while in the second definition both coordinates are irrationnals.

$\mathbb R^2 \setminus \mathbb Q^2=(\mathbb R \setminus \mathbb Q)^2\quad\cup\quad(\mathbb R \setminus \mathbb Q)\times\mathbb Q\quad\cup\quad\mathbb Q\times(\mathbb R \setminus \mathbb Q)$.

But we have $\begin{cases} (\mathbb R \setminus \mathbb Q)^2\cup(\mathbb R \setminus \mathbb Q)\times\mathbb Q=(\mathbb R \setminus \mathbb Q)\times\mathbb R\quad\mathrm{a\ continuous\ vertical\ path}\\ (\mathbb R \setminus \mathbb Q)^2\cup\mathbb Q\times(\mathbb R \setminus \mathbb Q)=\mathbb R\times(\mathbb R \setminus \mathbb Q)\quad\mathrm{a\ continuous\ horizontal\ path} \end{cases}$

And this is precisely along these two sets of mixed coordinates that we can build continuous paths to connect points of $\mathbb R^2 \setminus \mathbb Q^2$.

For $(\mathbb R \setminus \mathbb Q)^2$ your construction of a positive semi-plane and a negative semi-plane works and this set is effectively disconnected.

zwim
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